Question

The elementary reversible reaction:

2A <----> B

is carried out isothermally and isobarically in a flow reactor where pure A is fed at a concentration of 4.0 mol/dm3. If the equilibrium conversion is found to be 60%. What is the equilibrium constant, Kc if the reaction is a gas phase reaction?

Answer 1

`Kc=0.9375`

Step-by-step explanation:

Hello,

In this case, considering that A's initial concentration is 4.0mol/dm³, considering the equilibrium conversion, one obtains for the equilibrium:

`C_A^(eq)=4.0mol/dm^3*(1-x_(eq))=4.0mol/dm^3*(1-0.6)=1.6mol/dm^3\\C_B^(eq)=4.0mol/dm^3*x_(eq)=2.4mol/dm^3`

Thus, the equilibrium constant turns out:

`Kc=(C_B)/(C_A^2) =(2.4)/(1.6^2)=0.9375`

Best regards.